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Zolol [24]
3 years ago
5

18 Select the correct answer.

Computers and Technology
1 answer:
Elden [556K]3 years ago
6 0

A string for sure, so answer B.

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Compared with traditional methods, the entire rapid application development (RAD) process is expanded and, as a result, the new
tensa zangetsu [6.8K]

Answer:

<u>False</u>

Explanation:

Note, the Rapid Application Development (RAD) software development approaches are noteworthy not for its expanded design details but for <u>its simplicity of the software development process.</u>

In other words, the RAD approaches while taking user preferences into the software development process, focuses on making the software design process more flexible, such as by employing the use of prototypes, and less unnecessary details.

5 0
2 years ago
attackers typically use ack scans to get past a firewall or other filtering device. how does the process of an ack scan work to
bogdanovich [222]

Attackers frequently use ACK scans to circumvent a firewall or other filtering tools. During a NULL scan, all packet flags are enabled. The most recent versions of Nessus Server and Client are compatible with Windows, Mac OS X, FreeBSD, and the vast majority of Linux variants.

<h3>What is ack scan ?</h3>
  • ACK scans are used to identify hosts or ports that have been blocked or are resistant to other types of scanning. An attacker uses TCP ACK segments to learn about firewall or ACL configuration.
  • Attackers probe our router or send unsolicited SYN, ACK, and FIN requests to specific UDP/TCP ports.
  • TCP ACK Scan sends an ACK message to the target port to determine whether or not it is filtered.
  • On unfiltered ports, a RST reply packet will be sent for both open and closed ports. Filtered ports will either generate no response or generate an ICMP reply packet with an unreachable destination.
  • The TCP ACK scanning technique attempts to determine whether a port is filtered by using packets with the ACK flag set.

To learn more about ask scan refer to:

brainly.com/question/13055134

#SPJ4

3 0
1 year ago
Thanks for help evreybody
ladessa [460]

Answer:

np

Explanation:

7 0
3 years ago
Read 2 more answers
Write a method called findNames that meets the following specs: It takes two arguments: a list of strings, allNames, and a strin
Anna11 [10]

Answer:

public class Solution {

   public static void main(String args[]) {

       String[] allNames = new String[]{"Bob Smith", "Elroy Jetson", "Christina Johnson", "Rachael Baker", "cHRis", "Chris Conly"};

       String searchString = "cHRis";

       

       findNames(allNames, searchString);

   }

   

   public static void findNames(String[] listOfName, String nameToFind){

       ArrayList<String> resultName = new ArrayList<String>();

       

       for(String name : listOfName){

           if (name.toLowerCase().contains(nameToFind.toLowerCase())){

               resultName.add(name);

           }

       }

       

       for(String result : resultName){

           System.out.println(result);

       }

   }

}

Explanation:

The class was created called Solution. The second line define the main function. Inside the main function; we initialized and assign an array called allNames to hold the list of all name. Then a String called searchString was also defined. The searchString is the string to search for in each element of allNames array. The two variables (allNames and searchString) are passed as argument to the findNames method when it is called.

The method findNames is defined and it accept two parameters, an array containing list of names and the name to search for.

Inside the findNames method, we initialized and assigned an ArrayList called resultName. The resultName variable is to hold list of element found that contain the searchString.

The first for-loop goes through the elements in the allNames array and compare it with the searchString. If any element is found containing the searchString; it is added to the resultName variable.

The second for-loop goes through the elements of the resultName array and output it. The output is empty if no element was found added to the resultName variable.

During comparison, the both string were converted to lower case before the comparison because same lowercase character does not equal same uppercase character. For instance 'A' is not same as 'a'.

8 0
3 years ago
Convert to binary 140
Ber [7]

00110001 00110100 00110000

5 0
3 years ago
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